R Standardized Precipitation Index .nc file

Question

I'm trying to calculate the SPI from CHIRPS monthly mean precipitation data, because it's too large I cut it down to my area of interest and here it is: https://www.dropbox.com/s/jpwcg8j5bdc5gq6/chirps_mensual_v1.nc?dl=0 I did this to open it:

require(utils)
require(colorRamps)
require(RNetCDF)
require(rasterVis)
require(rgdal)
library(ncdf4)
library(raster)


datos2 <- nc_open("Datos/chirps_mensual_v1.nc")
ppt_array <- ncvar_get(datos2, "precip")

#I'm only taking complete years so I took out two months from 2018

ppt_mes <- ppt_array[ , ,1:444]

I know there is a SPI library but I don't know how should I format the data in order to use it. So I tried to do it without the function by fitting the gamma distribution but I dont' know how to do it for this data base.

Does anyone know how to calculate SPI either with the function or by fitting the distribution?

Answer 1

At the end I made it by using the SPI library, the result will be a value for each month in each grid point, if you want to calculate the value over a specific area I made that too but I can share it if you want it too:

Also, this one I made it using CRU data but you can adjust it:

#spei cru 1x1
rm(list=ls(all=TRUE)); dev.off()

require(utils)
require(RNetCDF)
require(rasterVis)
require(rgdal)
library(ncdf4)
require(SPEI)

########################################################################################################


prec <- open.nc("pre_mensual.nc")

lon <- length(var.get.nc(prec, "lon"))
lat <- length(var.get.nc(prec, "lat"))
lon1 <- var.get.nc(prec, "lon")
lat1 <- var.get.nc(prec, "lat")
ppt  <- var.get.nc(prec, "pre") 
ppt  <- ppt[ , ,109:564] #31 18 456 (1980-2017)
anio = 456/12

###########################################################################################################
#Reshape data 

precip <- sapply(1:dim(ppt)[3], function(x)t(ppt[,,x]))

############################################################################################################
#This is for SPI-6, you can use either of them

spi_6 <- array(list(),(lon*lat))

for (i in 1:(lon*lat)) {
  spi_6[[i]] <- spi(precip[i,], scale=6, na.rm=TRUE)
}
#############################################################################################################
#Go back to an array form

sapply(spi_6, '[[',2 )->matriz_ppt 
ppt_6 <- array(aperm(matriz_ppt, c(2,1),c(37,63,456)));spi_c <- array(t(ppt_6), dim=c(37,63,456))
#############################################################################################################
    #Save to netcdf

for(i in 1:456) { 
  nam <- paste("SPI", i, sep = "")
  assign(nam,raster((spi_c[ , ,i]), xmn=min(lon1), xmx=max(lon1), ymn=min(lat1), ymx=max(lat1), crs=CRS("+proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs+ towgs84=0,0,0")) )
}

gpcc_spi <- stack(mget(paste0("SPI", 1:456)))

outfile <- "spi6_cru_1980_2017.nc"
crs(gpcc_spi) <- "+proj=longlat +datum=WGS84 +no_defs +ellps=WGS84 +towgs84=0,0,0" 
writeRaster(gpcc_spi, outfile, overwrite=TRUE, format="CDF", varname="SPEI", varunit="units",longname="SPEI CRU", xname="lon", yname="lat")

It's not the most stylish way to calculate it but it does work. :)

EDIT: If you want to calculate the SPI/SPEI over an area this is what I did:

library(SPEI)
library(ncdf4)
library(raster)
#

pre_nc <- nc_open("pre_1971_2017_Vts4.nc")
pre <- ncvar_get(pre_nc, "pre")
pre <- pre[, , 109:564] #This is for the time I'm interested in
lats <- ncvar_get(pre_nc, "lat")
lons <- ncvar_get(pre_nc, "lon")
times <- 0:467

# Read mask

#This is a mask you need to create that adjusts to your region of interest
#It consist of a matrix of 0's and 1's, the 1's are placed in the area
#you are interested in

mask1 <- nc_open("cuenca_IV_CDO_05_final.nc")
m1 <- ncvar_get(mask1, "Band1")
m1[m1 == 0] <- NA
#
# Apply mask to data
#
pre1 <- array(NA, dim=dim(pre))

#
for(lon in 1:length(lons)){
  for(lat in 1:length(lats)){
    pre1[lon,lat,] <- pre[lon,lat,]*m1[lon,lat]
  } 
}

#
# Mean over the area of interest
#
mean_pre1 <- apply(pre1,c(3),mean, na.rm=TRUE)

# Calculate SPI/SPEI

spi1 <- matrix(data= NA, nrow = 456, ncol = 48)
for (i in 1:48) {
  spi1[,i] <- spi(data=ts(mean_pre1,freq=12),scale= i)$fitted
}

#This calculates SPI/SPEI-1 to SPI/SPEI-48, you can change it
# Save
#
write.table(spi1,'spi_1980_2017.csv',sep=';',row.names=FALSE)

Answer 2

I don't think the SPI package is doing what you (or anyone) thinks it is doing. If you use debug(spi) and step through the code, you'll see that in one step it fits a empirical cumulative distribution function (with ecdf()) to the first two and last rows of data. Why the first two and last rows? I have no clue, but whoever wrote this package also used a for loop to do t() to a matrix. Not to mention that I think it should use a Gamma distribution or Pearson III distribution not ecdf() (according to Guttman, N.B. (1999) Accepting the standardized precipitation index: a calculation algorithm. JAWRA Journal of the American Water Resources Association, 35, 311–322.).